Simplify \( \frac{n!}{r!(n-r)!}+\frac{2(n+1)!}{r!(n-r+1)!} \)

Simplify the expression by following steps: - step0: Solution: \(\frac{n!}{r!\times \left(n-r\right)!}\) Calculate or simplify the expression \( \frac{2(n+1)!}{r!(n-r+1)!} \). Simplify the expression by following steps: - step0: Solution: \(\frac{2\left(n+1\right)!}{r!\times \left(n-r+1\right)!}\) Calculate or simplify the expression \( \frac{2(n+1)!}{r!(n-r+1)!} \). Simplify the expression by following steps: - step0: Solution: \(\frac{2\left(n+1\right)!}{r!\times \left(n-r+1\right)!}\) Calculate or simplify the expression \( \frac{2(n+1)\cdot n!}{r!(n-r+1)!} \). Simplify the expression by following steps: - step0: Solution: \(\frac{2\left(n+1\right)\times n!}{r!\times \left(n-r+1\right)!}\) - step1: Multiply the terms: \(\frac{2\left(n+1\right)!}{r!\times \left(n-r+1\right)!}\) To simplify the expression \[ \frac{n!}{r!(n-r)!} + \frac{2(n+1)!}{r!(n-r+1)!} \] we can break it down into two parts. 1. The first term is \[ \frac{n!}{r!(n-r)!} \] This is already in its simplest form, representing the binomial coefficient \( \binom{n}{r} \). 2. The second term is \[ \frac{2(n+1)!}{r!(n-r+1)!} \] We can rewrite \( (n+1)! \) as \( (n+1) \cdot n! \), so we have: \[ \frac{2(n+1) \cdot n!}{r!(n-r+1)!} \] Now, we can combine both terms: \[ \frac{n!}{r!(n-r)!} + \frac{2(n+1) \cdot n!}{r!(n-r+1)!} \] To combine these fractions, we need a common denominator. The common denominator will be \( r!(n-r+1)! \). Thus, we rewrite the first term: \[ \frac{n! \cdot (n-r+1)}{r!(n-r+1)!} \] Now we can combine the two fractions: \[ \frac{n!(n-r+1) + 2(n+1)n!}{r!(n-r+1)!} \] Factoring out \( n! \): \[ \frac{n! \left( (n-r+1) + 2(n+1) \right)}{r!(n-r+1)!} \] Now simplifying the expression inside the parentheses: \[ (n-r+1) + 2(n+1) = n - r + 1 + 2n + 2 = 3n - r + 3 \] Thus, the final simplified expression is: \[ \frac{n!(3n - r + 3)}{r!(n-r+1)!} \] This is the simplified form of the original expression.